我需要的是通過另一個變數動態地（在回圈內）增加一個變數的值。這是我的代碼：

while read  Plate City Town Village Area Population; do
    echo "$Plate $City $Town $Village $Area $Population"
    
     printf "\n$totalNumberOfTowns\n " 
     totalNumberOfTowns=$totalNumberOfTowns $Town
     totalNumberOfVillages=$totalNumberOfVillages $Village

done < "cities.txt"

運行代碼時得到的結果有點不正確。

我不明白為什么 $totalNumberOfTowns 變數中有“Town”。另一個有趣的部分是它還進行計算（它將第三個引數相加，即該城市的城鎮數量。對于 ADANA，它是 15，對于 ADIYAMAN，它是 9，依此類推。）。

編輯：

city.txt 的第一行是：

板市鎮村區人口

看截圖：

Edit-2: # 我嘗試過這些不同的語法：

let "totalNumberOfTowns =$Town" #1 I tried these different syntaxes
totalNumberOfTowns=$((expr $totalNumberOfTowns   $Town)) #2
totalNumberOfTowns=$((totalNumberOfTowns   Town)) #3

然而，他們每個人都給了我一些錯誤。

以下是示例 city.txt 和示例代碼：

平板城市城鎮村莊地區人口1 Adana 15 508 14030 2258718 2 Adiyaman 9 420 7614 632459 3 afyonkarahiSar 18 395 14230 736912 4A?ri8 5666666666611376 11376 535435 535435 5 AMASYA 7 352 5520 333 333 33 3334 6 AN ANINALY 33 3334 6 AN AN ANANAKARAKARAKARARA 40 166 5196 15462452

代碼

#!/usr/bin/env bash

average=0
numberOfCities=80

declare -i totalNumberOfTowns=0 totalNumberOfVillages=0
arrayWithOut=""

while read  Plate City Town Village Area Population; do
    echo "$Plate $City $Town $Village $Area $Population"

    printf "\n$totalNumberOfTowns\n " 
      totalNumberOfTowns =$Town
     totalNumberOfVillages =$Village

done < "cities.txt"

uj5u.com熱心網友回復：

您必須忽略第一行。

#!/usr/bin/env bash

average=0
numberOfCities=80

declare -i totalNumberOfTowns=0 totalNumberOfVillages=0
arrayWithOut=""

{
read -r first_line
while read -r Plate City Town Village Area Population; do
    echo "$Plate $City $Town $Village $Area $Population"

    printf "\n$totalNumberOfTowns\n " 
      totalNumberOfTowns =$Town
     totalNumberOfVillages =$Village

done 
} < "cities.txt"

uj5u.com熱心網友回復：

您還可以使用具有整數屬性的變數：

#!/usr/bin/env bash

declare -i totalNumberOfTowns=0 totalNumberOfVillages=0

while read Plate City Town Village Area Population; do
    printf "%s %s %s %s %s %s\n" "$Plate" "$City" "$Town" "$Village" "$Area" "$Population"
    totalNumberOfTowns =$Town
    totalNumberOfVillages =$Village
done < "cities.txt"

printf "Total towns: %d\nTotal villages: %d\n" "$totalNumberOfTowns" "$totalNumberOfVillages"

uj5u.com熱心網友回復：

我什至知道三種在 bash 中計算的方法：

命令 expr（運算子： 、-、*、/、%、(、)；和邏輯運算子：<、<=、==、!=、>=、>）

expr 2 * 3（運算子前的空格和''符號很重要）

或者

a=5；a=$(expr $a 1)

語法 (( <expression> )) - C 語言中已知的所有運算子

a=0；a=$((a 1))

命令讓

x=0；讓“x = 2”

uj5u.com熱心網友回復：

例如，只需將運算式放在雙括號中$((your_expression))

while read  Plate City Town Village Area Population; do
    echo "$Plate $City $Town $Village $Area $Population"
    
     printf "\n$totalNumberOfTowns\n " 
     totalNumberOfTowns=$((totalNumberOfTowns   Town))
     totalNumberOfVillages=$((totalNumberOfVillages   Village))

done < "cities.txt"

uj5u.com熱心網友回復：

一個可行的（而且很可能更快）的awk替代方案。

awk '
NR==1{print} 
NR>1{a =$3;b =$4;print}
END{
  print "Total number of towns:",a
  print "Total number of villages:",b
}' cities.txt

標籤：linux重击整数

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